Research website of Vyacheslav Gorchilin
2019-02-25
Calculation of energy of the bias current
Actually, the main thing is the basic principle mentioned in the previous section, it now remains the most simple — all that counting. But it is also an important component of the future device, because mathematics is our link with reality! You can collect a number of schemes, but not to the desired effect just because a key item was not designed for the required time parameter.
All our further calculations will be based on certain assumptions, without which the equations will be prohibitively bulky and not the fact that will be solved analytically. We assume that the condenser, through which flows the bias current, ideal and has no losses. The same applies to other elements of the schema that is presented in figure 2 in the first part of this work. An inductor located between the plates of a capacitor has no internal resistance and does not prevent the bias current that, if time is not so far away from reality.
Our task will be based on two connected among themselves events. First key SW1 is closed and on the plates of the capacitor C1 begins to increase potential, which constitutes the bias current, and he, in turn, generates a magnetic field $$B$$. Its something we have to find in the first step, with the help of the law of ampere-Maxwell and the equations (1.3).
In the second step is the energy of the magnetic momentum and, assuming that it is utilized with a coil L1 and a fully rooted in the active load R1, is compared with the time expended. Thus, we calculated the energy and the power in the load, and find the efficiency of the entire scheme. It will be announced at the beginning of this engineering formula.
Step # 1
First, let's consider the right part of equation (1.3), which is responsible for the conversion of electrical to magnetic fields. Because we believe that the flow of electric induction are uniformly distributed on the surface $$S$$ and because $$D = \sigma S$$, where $$\sigma$$ is the charge density, we find this integral: $\int \limits _{S}{\frac {\partial \mathbf {D} }{\partial t}}\, \mathbf {dS} = \frac{d (\sigma S)}{d t} = \frac{d q}{d t} = C \frac{d U}{d t} \qquad (2.1)$ In it: $$q$$ is the charge on the capacitor C1, $$C$$ is its capacity, $$U$$ is the voltage on this capacitor, changing the law $$U = U(t)$$. As you can see, despite the initial complexity of the integral, under certain assumptions, it has become a fairly simple formula with known parameters.
The left part of equation (1.3), in fact, perceive the generated magnetic field and it is a tension along a closed path $$\ell$$. If you look at the line of this field (Fig. 1), it becomes evident that their length is simply equal to the length of the circle: $$\ell=2\pi r$$. Then this integral also turns into clear and calculated the formula: $\oint \limits _{\ell}\mathbf {H}\, \mathbf {dl} = H\,2\pi r \qquad (2.2)$
But we are interested in the average radius of $$r_m$$, which will be the average length of the magnetic line of the coil pickup, which is just: $$r_0=2 r_m$$, where $$r_0$$ is the radius of the plates of the capacitor. Equate (2.1) and (2.2) $H = \frac{C}{\pi r_0} \frac{d U}{d t} \qquad (2.3)$ and, remembering the formula for flat circular capacitor, $C = \varepsilon\varepsilon_0 {\pi r_0^2 \over d} \qquad (2.4)$ substitute it in (2.3) is $H = \varepsilon\varepsilon_0 {r_0 \over d} \frac{d U}{d t} \qquad (2.5)$ where: $$d$$ — distance between the plates of the capacitor. We got a formula that can calculate the magnetic field strength obtained from the electric conversion method. Her well-observed pattern: the higher the amplitude and the slew rate of the electric field, the stronger will be the magnetic pulse. You can feel the physical vibration of the coil L1 (and reactors are described in the following sections) at a high enough specified parameters.
Step # 2
By itself, the field strength tells us nothing about its energy or power. But if you take to take another Maxwell's formula for the energy of the magnetic field [1], the situation becomes clear immediately: $W_2 = {\mu\mu_0\,H^2 \over 2} V \qquad (2.6)$ Here $$W_2$$ is the energy of the magnetic field and $$V$$ is the volume in which it is contained, and which is by the classic formula: $$V = \pi r_0^2\,d$$.
We assume that fully recycle all that energy using toroidal coil L1 and the load R1 (Fig. 2), so we can substitute in the formula (2.6) the field strength of (2.5): $W_2 = {\mu\mu_0\,(\varepsilon\varepsilon_0)^2 \over 2} {\pi r_0^3 \over d} \left(\frac{d U}{d t}\right)^2 \qquad (2.7)$ Thus, we found the energy that can obtained using the method of conversion from one edge of the first pulse generated by the key SW1 and the capacitor C1 (Fig. 2).
The energy required to charge C1, we assume a simple classical formula which does not depend on the way of charge: $W_1 = C\,U_0^2 \qquad (2.8)$ $$U_0$$ is the value of the voltage on the capacitor C1 at the time of the rupture of key SW1. At the same time we must not forget that the charge is spent two times more energy than is stored in the capacitor, so the multiplier 1/2 is absent here.
It remains to compare these energy and to get the value of growth efficiency: $K_{\eta2}= {W_2 \over W_1} \qquad (2.9)$ But to get the final formula will make another layout. Naturally, the increase of the potential on the capacitor plates will occur from zero up to some maximum value. Reflect this: $\frac{d U}{d t} = U_0 \frac{d F}{d t} \qquad (2.10)$ where: $$U_0$$ is the value of the voltage on the capacitor C1 at the time of the rupture of key SW1, $$F$$ is a function of rise of this voltage after the circuit of the key, which in General case depends on time: $$F=F(t)$$.
Then, substituting the formula (2.10) in (2.9), and instead of $$C$$ its value from (2.4) we obtain the final result: $K_{\eta2}= \mu\mu_0\,\varepsilon\varepsilon_0 {r_0 d \over 2} \left(\frac{d F}{d t}\right)^2 \qquad (2.11)$ now, we would like to draw your attention to the "speed characteristics" of the increase of the electric field in comparison with the speed of light. To do this, remember that $$\mu_0\varepsilon_0 = 1/c^2$$, where $$c$$ is the speed of light. And if, for simplicity, suppose we do the experiment in a vacuum and without ferromagnetic cores, and therefore the relative dielectric permittivity and magnetic permeability equal to one, we come to a very simple formula $K_{\eta2}= {r_0 d \over 2} \left(\frac{1}{c} \frac{d F}{d t}\right)^2 \qquad (2.12)$ which tells us that $$K_{\eta2} \gt 1$$, the rate of rise of voltage across the capacitor, in combination with the $$\sqrt{r_0 d}$$ should be comparable with the speed of light.
Example 1
Again, recall that between the plates of the capacitor C1 is a toroidal coil L1, the average radius which, for ease of calculation, assumed equal to half the radius of the plates of the capacitor. Its purpose is to utilizzare magnetic pulse in the load R1 (Fig. 2).
 Fig.3. Three graphs of the voltage rise on the capacitor C1 depending on the time — linear, exponential, and perfect
If we assume that the increase in the voltage on C1 occurs under the linear law (Fig. 3a) $U = U_0 \frac{t}{T_0}, \quad t \in [0, T_0] \qquad (2.13)$ where $$T_0$$ is the time of disconnection of key SW1, then the slew rate will be $\frac{d F}{d t} = {1 \over T_0} \qquad (2.14)$ and the increase in efficiency will be so: $K_{\eta2} = {r_0 d \over 2} \left(\frac{1}{c\,T_0}\right)^2 \qquad (2.15)$ In this example, we assume that $$\mu = \varepsilon = 1$$. Then in order to the efficiency of the scheme was greater than one, you must perform the following relation: $T_0 \lt \frac{1}{c} \sqrt{{r_0 d \over 2}} \qquad (2.16)$ for Example, if you take the radius of the capacitor plates and the distance between them is 30cm, the edge of the pulse should be less than 0.7 NS.
Example No. 2
If you take the exponential, a more realistic dependence of rise of voltage across C1 (Fig. 3b), $U = 1.59\,U_0 \left(1 - e^{-t/T_0} \right), \quad t \in [0, T_0] \qquad (2.17)$ then, by taking the first derivative to find the rate of growth: $\frac{d F}{d t} = 1.59 {e^{-t/T_0} \over T_0} \qquad (2.18)$ In the time $$t = T_0$$ this speed will be approximately equal to $\frac{d F}{d t} = {0.59 \over T_0} \qquad (2.19)$ and growth efficiency would be: $K_{\eta2} = 0.35 {r_0 d \over 2} \left(\frac{1}{c\,T_0}\right)^2 \qquad (2.20)$ If we now apply the algorithm and data according to the formula (2.16), the condition is super $$K_{\eta2}$$ will be performed if time front impulse will be less than 0.4 NS. This result would be more consistent with reality.
Based on these examples it is possible to imagine the curve of rise of voltage across the capacitor will be more perfect. Figure 3c depicts such a schedule. In real structures this can be achieved by including in the charging circuit of small inductance with low capacitance, however, we should not forget that it will increase the time $$T_0$$.

Despite the fact that in this part of the work is clearly visible approach to the method of calculating the switching circuits, based on the bias currents and the transformation of fields, the design of the device is not perfect. In the next part we will offer a more realistic circuitry and change its calculation.

The materials used
1. Electromagnetic induction. The energy of the magnetic field. Formula 27.
2. Experiments on the detection and study of the displacement currents in a vacuum.